1、自動化學院隨機過程課程設計學 號： S308040031專 業： 導航、制導與控制學生姓名：任課教師： 趙希人 教授2008年12月第一題、源程序：x=rand(1,2000) EX=mean(x)%均質檢驗DX=var(x) %方差檢驗subplot(2,1,1),hist(x,10);y=linspace(-10,10,21);for m= -10 :10 mAbs=abs(m); s=0; for n=1:2000-mAbs s=s+(x(n+mAbs)-EX)*(x(n)-EX); end y(m+11)=(1/(2000-mAbs)*s;endx2=-15:30/20:15;subp

2、lot(2,1,2),plot(x2,y);1.打印前50個數：Columns 1 through 120.3329 0.5189 0.5435 0.9038 0.2451 0.1324 0.0977 0.6975 0.3707 0.1985 0.5770 0.4111Columns 13 through 24 0.1570 0.7128 0.3445 0.6434 0.4711 0.3107 0.9610 0.2236 0.0607 0.2657 0.3861 0.3220Columns 25 through 36 0.3933 0.3022 0.7159 0.4312 0.5941 0.9

3、034 0.1390 0.1119 0.8127 0.1762 0.5175 0.0592Columns 37 through 48 0.4535 0.7152 0.1639 0.1661 0.2440 0.2296 0.9726 0.5498 0.7901 0.3500 0.2899 0.8619Columns 49 through 50 0.9937 0.3002 2.分布檢驗：3.均值檢驗：EX = 0.50384.方差檢驗：DX = 0.08275.計算相關函數：第二題、源程序：clear all;for n=1:2000xt(n)=normrnd(0,1);%產生2000個N(0,1

4、)分布的獨立序列endplot(xt),title('2000個N(0,1)分布的獨立序列');figurefor i=1:5 for j=1:10 sc(j,i)=xt(i-1)*5+j); end;end;disp(sc);EX=mean(xt)%求平均數并輸出DX=cov(xt)%求方差并輸出subplot(2,1,1);p=hist(xt,20);%將產生的2000個隨機數分為20組p=p/100; t=-2.85:0.3:2.85;%求概率密度bar (t,p,1) ,title('N(0,1)分布的獨立序列的直方圖');xlabel('x&#

5、39;);ylabel('f(x)');tx,i = xcov(xt,10);%取-10到10Tx=tx/2000;%求自相關函數x()subplot(2,1,2)plot(i,Tx, '.- '),title('自相關函數x()分布圖');xlabel('');ylabel('x()');1.打印前50個數：- 12 -0.3599 0.7562 1.0020 1.3747 0.00460.6307 1.1306 -0.4553 0.0294 -0.5836-0.6737 -0.8743 -0.0972 -1.6

6、200 0.5674 0.7104 -0.8068 0.7932 -0.3971 0.42000.6455 -0.0149 -1.1756 -0.0226 -0.36670.7562 1.0020 1.3747 0.0046 0.5132 1.1306 -0.4553 0.0294 -0.5836 1.6777 -0.8743 -0.0972 -1.6200 0.5674 -0.5122 -0.8068 0.7932 -0.3971 0.4200 0.4997 -0.0149 -1.1756 -0.0226 -0.3667 0.6410 程序運行產生的2000個N（0，1）分布的獨立序列：2.

7、分布檢驗：3.均值檢驗：EX = 0.00694.方差檢驗：DX = 0.99035.計算相關函數：第三題、源程序：y=normrnd(0,1,1,2000); x=zeros(1,2000);for k=2:2000 x(k)=y(k)+4*y(k-1);end;Ex=mean(x) %求EX（k）Ex2=mean(x.2) %EX.2(k)Dx=var(x) %DX(k)Bm=zeros(1,11); for i=0:10 xy=zeros(1,2000-i); for j=1:(2000-i) xy(j)=(x(j+i)-Ex)*(x(j)-Ex); end; Bm(i+1)=sum(x

8、y)/(2000-i); %Bx(m)end;a=Bm(:,11:-1:1);a=a(1:10),Bm;Bm=aq=-10:10;plot(q,Bm)1.求均值EX：Ex = 0.17672.求二階矩EX2：EX2 = 16.66883.求方差：Dx = 16.64594.BX（m）：Columns 1 through 12 0.5104 -0.1167 -0.4110 0.1765 0.4423 0.1237 0.2300 -0.3652 -0.4433 4.0652 16.6376 4.0652Columns 13 through 21 -0.4433 -0.3652 0.2300 0.1

9、237 0.4423 0.1765 -0.4110 -0.1167 0.5104第四題、源程序：y=normrnd(0,1,1,2000);x=zeros(1,2000);for k=2:2000 x(k)=y(k)-0.707*x(k-1);end;M=zeros(1,901);for k=101:1001 x(k)=x(k)+x(k-1); M(k-100)=x(k-1);end;Ex=x(k)/900 %EX(k)Ex2=sum(M.2)/900 %EX.2(k)Dx=Ex2-(Ex).2 %DX(k)Bm=zeros(1,11);for i=0:10 xy=zeros(1,1900-i

10、); for j=100:(2000-i) xy(j-99)=(x(j+i)-Ex)*(x(j)-Ex); end; Bm(i+1)=sum(xy)/1900 %Bx(m)end;a=Bm(:,11:-1:1);a=a(1:10),Bm;q=-10:10;plot(q,a) %Bx(m)二維分布曲線1.求EX：Ex = -0.00252.求Ex2：Ex2 = 39.81773.求DX：Dx = 39.81774.求Bx（m）：Bm = 19.8593 17.7436 19.0243 18.0245 18.4574 18.0159 18.0863 17.8381 17.7725 17.5919

11、17.4769第五題、 源程序：f='sin(x)' %畫出符號函數subplot(3,2,1);ezplot(f)subplot(3,2,2);ezplot(f)n=-20:20;y=sin(n*pi/2);subplot(3,2,3);k=-10:10;plot(k,sin(k*pi/2),'-rs','MarkerEdgeColor','k','MarkerFaceColor','g','MarkerSize',5)title('采樣信號(±10點)'

12、);subplot(3,2,4);k=-20:20;plot(k,sin(k*pi/2),'-rs','MarkerEdgeColor','k','MarkerFaceColor','g','MarkerSize',5)title('采樣信號(±20點)');D=0.05;z1=1;for t1=-5*pi:D:5*pi; s1=0; for m1=-10:10 s1=s1+y(m1+11)*sinc(1/pi)*(t1-pi*m1/2); %sinc函數內插恢復，重建信號

13、過程 end fa(z1)=s1;z1=z1+1;endsubplot(3,2,5)xlab1=linspace(-5*pi,5*pi,length(fa);plot(xlab1,fa);title('內插恢復信號(±10點)');z2=1;for t1=-10*pi:D:10*pi; s2=0; for m2=-20:20 s2=s2+y(m2+21)*sinc(1/pi)*(t1-pi*m2/2); %sinc函數內插恢復，重建信號過程 end fb(z2)=s2;z2=z2+1;endsubplot(3,2,6)xlab2=linspace(-10*pi,10*

14、pi,length(fb);plot(xlab2,fb);title('內插恢復信號(±20點)');運行結果：說明：圖中第一行兩圖為sin函數的符號函數示意圖，區間-22。圖中第二行兩圖為sin函數的采樣函數示意圖，左圖區間-55（±10點），右圖區間-1010（±20點）。圖中第三行兩圖為sin函數的恢復函數示意圖，右圖比左圖更加接近原函數，但是由于是有限采樣點恢復，并且，內插sin c函數本身是非因果的連續信號，物理上不可實現，故只能用大樣本離散值模擬，程序中取步長D=0.05。所以，可以看到兩圖在函數邊緣有較大過沖。這是由于sin c 函數

15、被截斷而引起的。第六題、源程序：A=zeros(40,21);k=zeros(1,20);for i=1:21A(1,i)=2*i-1;endfor j=1:20if mod(j,2)=1;A(2,j)=A(1,j+1);elseA(2,j)=0;end;end;k(1)=A(1,1)/A(2,1);for i=3:40if mod (i,2)=1;for j=(i+1)/2:2:20;A(i,j)=A(i-2,j);if j=20;A(i,j+1)=A(i-2,j+1)-A(i-1,j+1)*k(i-1)/2);elseA(i,j+1)=41;end;endelsefor j=i/2:2:2

16、0A(i,j)=A(i-1,j+1);end;k(i/2)=A(i-1,i/2)/A(i,i/2);end;end;disp(A)disp(k)1.奧斯特姆表：A =Columns 1 through 12 1.0000 3.0000 5.0000 7.0000 9.0000 11.0000 13.0000 15.0000 17.0000 19.0000 21.0000 23.00003.0000 0 7.0000 0 11.0000 0 15.0000 0 19.0000 0 23.0000 00 3.0000 2.6667 7.0000 5.3333 11.0000 8.0000 15.0

17、000 10.6667 19.0000 13.3333 23.00000 2.6667 0 5.3333 0 8.0000 0 10.6667 0 13.3333 0 16.00000 0 2.6667 1.0000 5.3333 2.0000 8.0000 3.0000 10.6667 4.0000 13.3333 5.00000 0 1.0000 0 2.0000 0 3.0000 0 4.0000 0 5.0000 00 0 0 1.0000 0 2.0000 0.0000 3.0000 0.0000 4.0000 0 5.00000 0 0 0 0 0.0000 0 0.0000 0

18、0 0 0.00000 0 0 0 0 -Inf 0.0000 -Inf 0.0000 NaN 0 -Inf0 0 0 0 -Inf 0 -Inf 0 NaN 0 -Inf 00 0 0 0 0 -Inf NaN -Inf NaN NaN NaN -Inf Columns 13 through 21 25.0000 27.0000 29.0000 31.0000 33.0000 35.0000 37.0000 39.0000 41.000027.0000 0 31.0000 0 35.0000 0 39.0000 0 016.0000 27.0000 18.6667 31.0000 21.33

19、33 35.0000 24.0000 39.0000 41.00000 18.6667 0 21.3333 0 24.0000 0 41.0000 016.0000 6.0000 18.6667 7.0000 21.3333 8.0000 24.0000 -7.1250 06.0000 0 7.0000 0 8.0000 0 -7.1250 0 00.0000 6.0000 0.0000 7.0000 0 8.0000 43.0000 -7.1250 41.00000 0.0000 0 0 0 43.0000 0 41.0000 00.0000 -Inf 0.0000 NaN 0 -Inf 4

20、3.0000 -Inf 0-Inf 0 NaN 0 -Inf 0 -Inf 0 0NaN -Inf NaN NaN NaN -Inf NaN -Inf 41.0000 Ak=0.3333, 1.1250, 2.6667, NaN, NaN, NaN, NaN ,NaN, NaN, NaN, NaN, NaNB =Columns 1 through 12 0 0 0 0 0 0 0 0 0 0 0 03.0000 0 7.0000 0 11.0000 0 15.0000 0 19.0000 0 23.0000 00 0 0 0 0 0 0 0 0 0 0 00 2.6667 0 5.3333 0

21、 8.0000 0 10.6667 0 13.3333 0 16.00000 0 0 0 0 0 0 0 0 0 0 00 0 1.0000 0 2.0000 0 3.0000 0 4.0000 0 5.0000 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0.0000 0 0.0000 0 0.0000 0 0.00000 0 0 0 NaN NaN NaN NaN NaN NaN NaN NaN0 0 0 0 -Inf 0 -Inf 0 NaN 0 -Inf 00 0 0 0 0 NaN NaN NaN NaN NaN NaN NaNColumns 13 thro

22、ugh 21 0 0 0 0 0 0 0 1.0000 027.0000 0 31.0000 0 35.0000 0 39.0000 0 00 0 0 0 0 0 0 1.0000 00 18.6667 0 21.3333 0 24.0000 0 41.0000 00 0 0 0 0 0 0 1.0000 06.0000 0 7.0000 0 8.0000 0 -7.1250 0 00 0 0 0 0 0 0 1.0000 00 0.0000 0 0 0 43.0000 0 41.0000 0NaN NaN NaN NaN NaN NaN NaN NaN NaN-Inf 0 NaN 0 -In

23、f 0 -Inf 0 0NaN NaN NaN NaN NaN NaN NaN NaN NaNBk=0,0,0, NaN, NaN NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN2.由奧斯特姆表可以看出系統不穩定。3.第七題、源程序：n=7;N=13;x=1 0.5 0.58 -0.01 -0.0121 0.00005 0.00006;y=1 0.55 0 0 0 0 0;%x=1 0.7 0.5 -0.3;%y=1 0.3 0.2 0.1; A=zeros(N,n); %分配內存空間B=zeros(N,n);a=zeros

24、(n-1,1);b=a;for i=1:n, %賦表的初始值，前兩行及a(1),b(1); A(1,i)=x(i); B(1,i)=y(i);end;for i=1:n, A(2,i)=A(1,n-i+1);end;for i=1:n, B(2,i)=A(2,i);end;a(1)=A(1,n)/A(2,n);b(1)=B(1,n)/B(2,n); for i=2:n, for j=1:n-i+1, A(2*i-1,j)=A(2*i-3,j)-A(2*i-2,j)*a(i-1); B(2*i-1,j)=B(2*i-3,j)-B(2*i-2,j)*b(i-1); end; if i=n, break; end; for j=1:n-i+1, A(2*i,j)=A(2*i-1,n-i+2-j); end; for j=1:n, B(2*i,j)=A(2*i,j); end; a(i)=A(2*i-1,n-i+1)/A(2*i,n-i+1); b(i)=B(2*i-1,n-i+1)/B(2*i,n-i+1);end;AB1.奧斯特姆表：A = 1.0000 0.5000 0.5800 -0.0100 -0.0121 0.0001 0.0001 0.00